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# OCR MEI Mechanics 4 (M4) Tweet

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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1. OCR MEI Mechanics 4 (M4)
Hi, this thread is for anyone taking the OCR MEI M4 exam to talk about the content and problems in the module, and to talk generally about maths of course
2. Re: OCR MEI Mechanics 4 (M4)
How are people feeling for the exam next week? I've still got to get my head around the "dynamics of a rotating body" chapter a little more but aside from that it should be good
3. Re: OCR MEI Mechanics 4 (M4)
im feeling pretty good about this one! I am actually sitting M1 and M2 at the same time
4. Re: OCR MEI Mechanics 4 (M4)
same, the dynamics of rotating bodies is the hardest one for me! feeling pretty good about it though, been getting As and A*s on past papers so far so you're doing M1, M2 and M4 in the same session? is that even allowed?
5. Re: OCR MEI Mechanics 4 (M4)
apparently you are allowed to sit up to 6 hours of exams on each day, which is just enough for me It felt really weird doing M3 first though, and man it was so boring sitting in M1 lessons; so easy it defies belief
6. Re: OCR MEI Mechanics 4 (M4)
(Original post by B Jack)
How are people feeling for the exam next week? I've still got to get my head around the "dynamics of a rotating body" chapter a little more but aside from that it should be good
When you have got your head around it, would you mind explaining it to me?
7. Re: OCR MEI Mechanics 4 (M4)
I've noticed with all the M4 past papers, for variable forces there has never been a question based largely on gravitational energy (GMm/r^2) or work done/ impulse of force integrals, I reckon it might be worth just making sure you're good with those parts of variable forces because all the other past papers have basically been just mdv/dt, mvdv/dx or Pv. It's only a suggestion but based on how the questions in the S4 paper were quite different from previous years I wouldn't be surprised if it were the same with M4 too..
8. Re: OCR MEI Mechanics 4 (M4)
Good luck everyone!!
9. Re: OCR MEI Mechanics 4 (M4)
Good luck to you too!
10. Re: OCR MEI Mechanics 4 (M4)
The point when you realise that actually, it may have been worth revising for M4... :/ I didn't get the last part of question 3 - how can you find a point of equilibrium when you don't actually know the value of lambda - if lambda>2mg, then lambda could be infinite, in which case there are no points of equilibrium for the given range of 2theta (though thinking about it I guess pi/3 would be stable... but then in another case when lambda isn't infinite it wouldn't be...)
Anyway, apart from that I don't think I did too badly... I hope :/
Last edited by 4ever_drifting; 01-06-2012 at 11:20.
11. Re: OCR MEI Mechanics 4 (M4)
For the last part of 3 you had to show that for dv/d(theta) = 0, either cos(theta) = 0 (leads to pi/2) or sin(theta) = lambda/(2lambda-2mg). Given lambda > 2mg, this fraction is always less than one, so you can arcsin it without worrying about domains, so there is a point of equilibrium at
theta = arcsin(lambda/(2lambda-2mg)). A bit of mucking about with the same inequality showed that the second derivative was always positive for lambda > 2mg, so it was stable for all lambda > 2mg. (pi/2 was unstable).
Unless that's all wrong. Wouldn't be that surprised...
12. Re: OCR MEI Mechanics 4 (M4)
How did you do the first part of question 2? I only got the answer they got when I ignored gravity.
13. Re: OCR MEI Mechanics 4 (M4)
(Original post by cwt255)
For the last part of 3 you had to show that for dv/d(theta) = 0, either cos(theta) = 0 (leads to pi/2) or sin(theta) = lambda/(2lambda-2mg). Given lambda > 2mg, this fraction is always less than one, so you can arcsin it without worrying about domains, so there is a point of equilibrium at
theta = arcsin(lambda/(2lambda-2mg)). A bit of mucking about with the same inequality showed that the second derivative was always positive for lambda > 2mg, so it was stable for all lambda > 2mg. (pi/2 was unstable).
Unless that's all wrong. Wouldn't be that surprised...
Ah I see... I was trying to find definite points of stability. Oh well, I can afford to get 0UMS in this exam so I'm not too worried.
14. Re: OCR MEI Mechanics 4 (M4)
(Original post by james22)
How did you do the first part of question 2? I only got the answer they got when I ignored gravity.
You had to consider that at the equilibrium point, the tension in the string isn't zero, it's mg (to balance the weight). So at time t, the total tension is mg + kx. So with acceleration downwards, m(dv/dt) = mg - (mg + kx) so mv(dv/dx) = -kx. Spent ages trying to work that one out, Q2 was all M3 stuff really...
15. Re: OCR MEI Mechanics 4 (M4)
Yup, should definitely have revised a little for this exam At least this is going to be the one module that doesn't count as all towards overall grades for me...
16. Re: OCR MEI Mechanics 4 (M4)
(Original post by Beth1234)
Yup, should definitely have revised a little for this exam At least this is going to be the one module that doesn't count as all towards overall grades for me...
Which modules aren't you doing? I get two "spares" at the end of it, probably M3 and something...
17. Re: OCR MEI Mechanics 4 (M4)
I thought the exam went well, the question 3 part iii took a while to find the nature of the equilibrium point that wasn't pi/2, but in the end I managed to show it was stable. Other than that I thought it was a really nice paper.
For question two where you had to find v in terms of x, how did you explain why the velocity was the negative root rather than the positive root? (v=-sqrt((k/m)((a^2)-(x^-2))) )
18. Re: OCR MEI Mechanics 4 (M4)
i am glad i dont need this module
19. Re: OCR MEI Mechanics 4 (M4)
(Original post by B Jack)
I thought the exam went well, the question 3 part iii took a while to find the nature of the equilibrium point that wasn't pi/2, but in the end I managed to show it was stable. Other than that I thought it was a really nice paper.
For question two where you had to find v in terms of x, how did you explain why the velocity was the negative root rather than the positive root? (v=-sqrt((k/m)((a^2)-(x^-2))) )
I just took the downwards direction as positive from the start, so when whatever it was was going upwards, velocity must have been negative.
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